--- title: "FS of 1 + sinω₀t + 2cosω₀t + cos(2ω₀t + π/4)" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:ctfs-coefficients-phase-mixed" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-coefficients-phase-mixed site: BCT Engineering --- # FS of 1 + sinω₀t + 2cosω₀t + cos(2ω₀t + π/4) Signals and Systems · Chapter 2: Fourier Series - Asked in: 2075 Baisakh, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-coefficients-phase-mixed - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question Find Fourier series coefficients and associated phase of the signal with fundamental frequency ω₀: x(t) = 1 + sin(ω₀t) + 2cos(ω₀t) + cos(2ω₀t + π/4) ## Short answer Coefficients: a0 = 1, a1 = 1 - j/2, a-1 = 1 + j/2, a2 = 0.5 e^{jπ/4}, a-2 = 0.5 e^{-jπ/4}, others zero. Magnitudes: |a0|=1, |a±1|=√5/2 ≈ 1.118, |a±2|=0.5. Phases: ∠a0=0°, ∠a±1 = ∓26.57°, ∠a±2 = ±45°. ## Answer Write each term using Euler's formula, with fundamental frequency $\omega_0$: $$ \begin{aligned} x(t) &= 1 + \frac{1}{2j}\left[e^{j\omega_0 t} - e^{-j\omega_0 t}\right] + \left[e^{j\omega_0 t} + e^{-j\omega_0 t}\right] + \frac{1}{2}\left[e^{j(2\omega_0 t + \pi/4)} + e^{-j(2\omega_0 t + \pi/4)}\right] \\ &= 1 + \left(1 + \frac{1}{2j}\right)e^{j\omega_0 t} + \left(1 - \frac{1}{2j}\right)e^{-j\omega_0 t} + \frac{1}{2}e^{j\pi/4}e^{j2\omega_0 t} + \frac{1}{2}e^{-j\pi/4}e^{-j2\omega_0 t} \end{aligned} $$ Comparing with $x(t) = \sum_k a_k e^{jk\omega_0 t}$ and using $\frac{1}{2j} = -\frac{j}{2}$: $$ a_0 = 1, \quad a_1 = 1 - \frac{j}{2}, \quad a_{-1} = 1 + \frac{j}{2}, \quad a_2 = \frac{1}{2}e^{j\pi/4} = \frac{\sqrt2}{4}(1 + j), \quad a_{-2} = \frac{1}{2}e^{-j\pi/4} $$ $a_k = 0$ for $|k| > 2$. #### Magnitude and phase $$ |a_{\pm1}| = \sqrt{1^2 + 0.5^2} = \frac{\sqrt5}{2} = 1.118, \qquad \angle a_{\pm1} = \mp\tan^{-1}(0.5) = \mp 26.57^\circ $$ | $k$ | $a_k$ | $\lvert a_k\rvert$ | $\angle a_k$ | |---|---|---|---| | $-2$ | $0.354 - j0.354$ | 0.5 | $-45^\circ$ ($-\pi/4$) | | $-1$ | $1 + j0.5$ | 1.118 | $+26.57^\circ$ (0.464 rad) | | 0 | 1 | 1 | $0^\circ$ | | 1 | $1 - j0.5$ | 1.118 | $-26.57^\circ$ ($-0.464$ rad) | | 2 | $0.354 + j0.354$ | 0.5 | $+45^\circ$ ($\pi/4$) | ```text |a_k| 1.118 | | 1 | | | 0.5 | | | | | -+--+--+--+--+--> k -2 -1 0 1 2 angle a_k (deg) 45 | 26.57 | | -+--+--o--+--+--> k -26.57 | | -45 | -2 -1 0 1 2 ``` As expected for a real signal, $a_{-k} = a_k^*$: the magnitude is even and the phase is odd. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the quickest way to find the coefficients of a sum of sinusoids at harmonics of w0? A. Expand each term with Euler's formula and compare with the synthesis sum B. Integrate each term over infinity C. Use Parseval's relation D. Square the signal first Answer: A. Writing each sinusoid as complex exponentials lets you read the a_k directly. (Key point: Method) 2. What is a_1 for sin(w0 t) + 2cos(w0 t)? A. 1 + j/2 B. 2 - j C. 1 - j/2 D. 1/2 - j Answer: C. a_1 = 1 (from 2cos) + 1/(2j) (from sin) = 1 - j/2. (Key point: a_1 value) 3. What are |a_1| and its angle? A. 1.5 and 0 B. 1.118 and +26.57 degrees C. 0.5 and -45 degrees D. 1.118 and -26.57 degrees Answer: D. |1 - j0.5| = sqrt(1.25) = 1.118 and the angle is -tan^-1(0.5) = -26.57 degrees. (Key point: Magnitude and phase) 4. What is a_2 for cos(2 w0 t + pi/4)? A. (1/2) e^{j pi/4} B. e^{j pi/4} C. (1/2) e^{-j pi/4} D. 1/2 Answer: A. cos splits into two halves; the one with e^{j2 w0 t} carries phase +pi/4 and amplitude 1/2. 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